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SUBGROUP | ABSTRACT ALGEBRA

Definition: Let G be a group under a binary operation * and H non empty subset of G. Then H is said to be a subgroup of G if H itself forms a group under the same binary operation * in G.

Theorem: A non empty subset H of a group G, is a subgroup of G if and only if a, b ∊ H ⟹ ab-1 ∊H

Proof: Let H be a subgroup of G. Then H itself forms a group under the operation in G. Hence by the closure and the inverse properties in H, a, b ∊H ⟹a, b-1 ∊H ⟹ab-1∊H

Conversely let a, b ∊H ⟹ab ∊H

Then a ∊H ⟹ a a-1 = e∊H, where e is the identity element of G. Also e, b∊H ⟹eb-1=b-1∊H .

Thus a, b∊H ⟹a(b-1)-1 = ab∊H .

Also assertively holds for H, as it does for G. It is observed that H satisfies the defining proprieties of a group and so is a subgroup of G.                  (Proved)

Theorem: Let G be a group and H a nonempty finite subset of G. Then G is a subgroup of G if and only if a, b ∊H ⟹ab ∊H

Proof: Let H be a subgroup of G. Then H must be closed under the operation in G, i.e., a, b ∊H ⟹ ab ∊H .

Conversely let a, b ∊H ⟹ ab ∊H.

Then for any a∊H, we have a2 = a . a ∊H, a3 = a2 a ∊H, a4 = a3 a ∊H, …, an ∊H, where n is any positive integer.

Since H is finite, the elements a, a2, a3,…, an … in H cannot all be distinct. This means there exist positive integers r, s (r > s)

Such that ar = as ——————-(i)

Applying the cancellation law in G, to (i)

We have ar-s = e, the identity element of G.

Since r – s>0, ar-s = e ∊ H.

Also r – s -1 ≥ 0 and since a0 = e   ∀ a ∊ G,

We have ar-s-1 ∊ H. Again, a ar-s-1 = ar-s =e

⟹ ar-s-1 is the inverse a in G.

⟹ ar-s-1 = a-1 ⟹ a-1 ∊ H.

Hence H is a subgroup of G.  (Proved)

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